Question and Answers Forum

All Questions      Topic List

Number Theory Questions

Previous in All Question      Next in All Question      

Previous in Number Theory      Next in Number Theory      

Question Number 5216 by Yozzii last updated on 01/May/16

Let p_j represent the j−th prime number. Now, define the number n whose decimal representation is written out in terms of p_j (j∈N) in the following way: n=0.p_1 p_2 p_3 p_4 p_5 ...p_j p_(j+1) p_(j+2) ... or n=0.(2)(3)(5)(7)(11)...(521)(523)(541)... ⇒n=0.235711...521523541... Prove or disprove that n is irrational.

$${Let}\:{p}_{{j}} \:{represent}\:{the}\:{j}−{th}\:{prime}\:{number}. \\ $$$${Now},\:{define}\:{the}\:{number}\:{n}\:{whose} \\ $$$${decimal}\:{representation}\:{is}\:{written}\:{out} \\ $$$${in}\:{terms}\:{of}\:{p}_{{j}} \:\left({j}\in\mathbb{N}\right)\:{in}\:{the}\:{following} \\ $$$${way}: \\ $$$${n}=\mathrm{0}.{p}_{\mathrm{1}} {p}_{\mathrm{2}} {p}_{\mathrm{3}} {p}_{\mathrm{4}} {p}_{\mathrm{5}} ...{p}_{{j}} {p}_{{j}+\mathrm{1}} {p}_{{j}+\mathrm{2}} ... \\ $$$${or}\:{n}=\mathrm{0}.\left(\mathrm{2}\right)\left(\mathrm{3}\right)\left(\mathrm{5}\right)\left(\mathrm{7}\right)\left(\mathrm{11}\right)...\left(\mathrm{521}\right)\left(\mathrm{523}\right)\left(\mathrm{541}\right)... \\ $$$$\Rightarrow{n}=\mathrm{0}.\mathrm{235711}...\mathrm{521523541}... \\ $$$${Prove}\:{or}\:{disprove}\:{that}\:{n}\:{is}\:{irrational}. \\ $$$$ \\ $$$$ \\ $$

Answered by FilupSmith last updated on 01/May/16

If we assume n is rational, ∴ n=(a/b), a,b∈Z (non reducible) p_j =jth prime ∴n=(p_1 /(10))+(p_2 /(100))+(p_3 /(1000))+...+(p_j /(10^j )) n=((100p_1 +10p_2 )/(1000))+(p_3 /(1000))+(p_4 /(10000))+... n=((100000p_1 +10000p_2 +1000p_3 )/(1000000))+(p_4 /(10000))+... n=((10^(j−1) p_1 +10^(j−2) p_2 +...)/(10^j )) usure if this is correct thus continuing we can always reduce the fraction. ∴ irrational ??? is this enough to prove??? i dont think so. Please check over this I am usure of the reliability of this proof. I am sure I could be wrong!

$$\mathrm{If}\:\mathrm{we}\:\mathrm{assume}\:{n}\:\mathrm{is}\:\mathrm{rational}, \\ $$$$\therefore\:{n}=\frac{{a}}{{b}},\:\:{a},{b}\in\mathbb{Z}\:\:\:\:\left(\mathrm{non}\:\mathrm{reducible}\right) \\ $$$${p}_{{j}} ={j}\mathrm{th}\:\mathrm{prime} \\ $$$$ \\ $$$$\therefore{n}=\frac{{p}_{\mathrm{1}} }{\mathrm{10}}+\frac{{p}_{\mathrm{2}} }{\mathrm{100}}+\frac{{p}_{\mathrm{3}} }{\mathrm{1000}}+...+\frac{{p}_{{j}} }{\mathrm{10}^{{j}} } \\ $$$${n}=\frac{\mathrm{100}{p}_{\mathrm{1}} +\mathrm{10}{p}_{\mathrm{2}} }{\mathrm{1000}}+\frac{{p}_{\mathrm{3}} }{\mathrm{1000}}+\frac{{p}_{\mathrm{4}} }{\mathrm{10000}}+... \\ $$$${n}=\frac{\mathrm{100000}{p}_{\mathrm{1}} +\mathrm{10000}{p}_{\mathrm{2}} +\mathrm{1000}{p}_{\mathrm{3}} }{\mathrm{1000000}}+\frac{{p}_{\mathrm{4}} }{\mathrm{10000}}+... \\ $$$${n}=\frac{\mathrm{10}^{{j}−\mathrm{1}} {p}_{\mathrm{1}} +\mathrm{10}^{{j}−\mathrm{2}} {p}_{\mathrm{2}} +...}{\mathrm{10}^{{j}} }\:\:\:\:\:\:\:{usure}\:{if}\:{this}\:{is}\:{correct} \\ $$$$\mathrm{thus}\:\mathrm{continuing}\:\mathrm{we}\:\mathrm{can}\:\mathrm{always}\:\mathrm{reduce} \\ $$$$\mathrm{the}\:\mathrm{fraction}. \\ $$$$ \\ $$$$\therefore\:\mathrm{irrational}\:\:\:\:\:\:\:???\:\mathrm{is}\:\mathrm{this}\:\mathrm{enough}\:\mathrm{to}\:\mathrm{prove}??? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{i}\:\mathrm{dont}\:\mathrm{think}\:\mathrm{so}. \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{check}\:\mathrm{over}\:\mathrm{this}\:\mathrm{I}\:\mathrm{am}\:\mathrm{usure}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{reliability}\:\mathrm{of}\:\mathrm{this}\:\mathrm{proof}.\:\mathrm{I}\:\mathrm{am}\:\mathrm{sure}\:\mathrm{I}\: \\ $$$$\mathrm{could}\:\mathrm{be}\:\mathrm{wrong}! \\ $$

Commented by Rasheed Soomro last updated on 03/May/16

If p_(i ) consist of n digits then each digit has its own denominator, not (p_i /(10^i )) . For example place value of 11 in given number is not equal to ((11)/(10^5 )) but is equal to (1/(10^5 ))+(1/(10^6 )) .

$${If}\:{p}_{{i}\:} {consist}\:{of}\:{n}\:{digits}\:{then}\:{each}\:{digit} \\ $$$${has}\:{its}\:{own}\:{denominator},\:{not}\:\frac{{p}_{{i}} }{\mathrm{10}^{{i}} }\:. \\ $$$${For}\:{example}\:{place}\:{value}\:{of}\:\mathrm{11}\:{in}\:{given}\:{number} \\ $$$${is}\:{not}\:{equal}\:{to}\:\frac{\mathrm{11}}{\mathrm{10}^{\mathrm{5}} }\:{but}\:{is}\:{equal}\:{to}\:\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{10}^{\mathrm{6}} }\:. \\ $$

Commented by FilupSmith last updated on 04/May/16

ahh I see. My mistake

$$\mathrm{ahh}\:\mathrm{I}\:\mathrm{see}.\:\mathrm{My}\:\mathrm{mistake} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com